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Let $R$ be a commutative ring, and let $M$ be a simple $R$-module. Prove $\text{End}_R(M)$ is a division algebra.

Attempt: Since $M$ is simple for any $\phi:M \rightarrow M$, $M/\{0\}\cong \phi(M)=M$ by the first isomorphism theorem. Define a ring homomorphism $\phi:R \rightarrow \text{End}_R(M)$ such that $\phi(R) \subset Z(\text{End}_R(M))$.

Since $M$ is simple, all $\alpha \in \text{End}_R(M)$ must be bijective. So it is a division ring. Since $M$ is an $R$ module, the action of $R$ is defined by a ring homomorphism $$\sigma:R \rightarrow \text{End}_{Ab}(M)$$ So should I just define $\phi(r)=\sigma(r)$ for each $r$? Would $\sigma(r)$ commute with each element in $\text{End}_R(M)$ since $R$ is commutative? How do I know each $\sigma(r) \in \text{End}_R(M)$?

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  • $\begingroup$ I think that $\operatorname{End}_R(M)$ is an associative $R$-algebra is obvious. So only the bijection part has to be proven. But you are right, it is more a mention than a proof. $\endgroup$ Oct 13 at 21:40
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There is an obvious map

$$\Phi: R \to \operatorname{End}_R(M)$$ defined by $$\Phi(r)(m) = rm.$$

Clearly $\Phi(r)$ is central for all $r \in R$, because if $f \in \operatorname{End}_R(M)$, then for all $m\in M$ $$(f\Phi(r))(m) = f(rm) = rf(m)= \Phi(r)(f(m)) = (\Phi(r)f)(m)$$ and thus $\Phi(r) \in Z(\operatorname{End}_R(M))$.

Using the ring homomorphism $\Phi: R \to Z(\operatorname{End}_R(M))$, the ring $\operatorname{End}_R(M)$ becomes an associative $R$-algebra.

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You are seeing the concept of $R$-algebra incorrectly. I think the following equivalence will be helpful for you.

For a (not necessarily commutative) ring $A$ the following statements are equivalent:

  • There is a ring homomorphism $R \to A$ whose image lies in $Z(A)$.
  • There is an $R$-module structure on $A$ such that $$(rx)y = r(xy) \quad\&\quad x(ry) = r(xy) \tag{1}$$ for all $r \in R$ and $x,y \in A$ (equivalently, the product $A \times A \to A,(x,y) \mapsto xy$ is $R$-bilinear).

Indeed, if there is a ring homomorphism $\nu : R \to A$, note that $A$ (being an $A$-module) gives us a ring homomorphism $\mu : A \to \operatorname{End}_\textsf{Ab}(A)$, and then $\mu \circ \nu$ yields an $R$-module structure to $A$. Explicitly, for $r \in R$ and $a \in A$, $$ra := \nu(r)a$$ (the product in $A$ of $\nu(r)$ with $a$). The equality $(1)$ is true if the image of $\nu$ lies in $Z(A)$.

Conversely, if there is an $R$-module structure on $A$ such that $(1)$ holds, then for each $r \in R$ and $a \in A$, $$(r1_A)a = r(1_Aa) = ra \quad\&\quad a(r1_A) = r(a1_A) = ra$$ so $(r1_A)a = a(r1_A)$. Hence, the image of the map $\_1_A : R \to A$, $r \mapsto r1_A$ is inside of $Z(A)$. Finally, from $(1)$ and the commutativity of $R$ one can also deduce that $(r_11_A)(r_21_A) = (r_1r_2)1_A$ for all $r_1,r_2 \in R$, meaning that $\_1_A$ is a ring homomorphism.


Thus, once is clear how $\operatorname{End}_R(M)$ is an $R$-algebra, it remains to be seen how the simplicity of $M$ implies that each nonzero $f \in \operatorname{End}_R(M)$ is invertible, but this is easy because $f \neq 0$ implies that $\ker f \neq M$ and $\operatorname{im} f \neq \{0\}$.

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